Tangents

To construct the tangent to a curve at a certain point A, you draw a line that follows the general direction of the curve at that point. An example of this can be seen below.

A concave up graph. A point (2, 2000) is marked by an x on the curved line. A tangent passes the curve at point A, and through the x axis at 1.0

Once the tangent is found you can use it to find the gradient of the graph by using the following formula:

\text{Gradient to the curve =}~\frac {y_2-y_1} {x_2-x_1}

where ({x_1,~y_1}) and ({x_2,~y_2}) are any two points on the tangent to the curve.

Question

Estimate the gradient to the curve in the graph below at point A.

A concave up graph. A point (2, 2000) is marked by an x on the curved line. A point A sits lower down the curved line

First draw the tangent at the point given.

A concave up graph. A tangent passes the curve at point A. A vertical line has been drawn from the tangent to the x axis at 2.5, and along the x axis to where the tangent passes through it at 1.0

Select any two points on the tangent. The coordinates that we are using are (1, 0) and (2.5, 2000). Then use the formula below:

\text{Gradient to the curve =}~\frac {y_2-y_1} {x_2-x_1}

\frac {2000~-~0} {2.5~-~1}~=~\frac {2000} {1.5}~=~1333.33

Positive and negative gradients

It is useful to remember that all lines and curves that slope upwards have a positive gradient.

Passengers on a rollercoaster travelling upwards. A line drawn under the cars represents the gradient. Because it is sloping upwards it has a positive gradient

All lines and curves that slope downwards have a negative gradient.

Passengers on a rollercoaster travelling downwards. A line drawn under the cars represents the gradient. Because it is sloping downwards it has a negative gradient

Example

We want to find the gradient of the curve at \text{x = -2}.

A graph with a concave up curve labelled 'y = x squared'

First draw the tangent at \text{x = -2}.

A graph with a concave up curve labelled 'y = x squared'. A tangent has been drawn next to the curve, passing through coordinates (-4, 9) and (0, -3)

Select any two points on the tangent. The coordinates that we are using are (-4, 9) and (0, -3).

\text{Gradient to the curve =}~\frac {y_2-y_1} {x_2-x_1}

\frac {y_2-y_1} {x_2-x_1} where ({x_1,~y_1})~=~({-4},{~9}) and ({x_2,~y_2})~=~({0},~{-3}) are any two points on the tangent to the curve.

Gradient to the curve = \frac {-3~-9} {0~-~(-4)}~=~\frac {-12} {4}~=~{-3}

Distance-Time graphs

If you are finding the gradient to the curve of a distance-time graph then you are calculating the velocity that the object is moving at that particular time.

Question

The following graph shows the car journey from Chelsea’s house to her mother’s house. Estimate the velocity of the car at \text{t = 6.5 s}. The tangent has been drawn for you.

A concave up graph. The x axis is labelled 'time', the y axis is labelled distance'. A tangent has been drawn next to the curve

\text{Gradient to the curve =}~\frac {y_2-y_1} {x_2-x_1}

\frac {140~-~20} {9~-~4}~=~\frac {120} {5}~=~24~ \text{m/s}^{2}

Velocity-Time graphs

Similarly, if you are finding the gradient to the curve of a velocity-time graph, then you are calculating the acceleration of the object at that particular time.